REAL NUMBERS | Irrational Numbers | Ex -2

Using Euclid's Division Lemma to Show: The Cube of Any Positive Integer is of the Form 9m, 9m + 1, or 9m + 8

Question:

Use Euclid's Division Lemma to prove that the cube of any positive integer is of the form \(9m\), \(9m + 1\), or \(9m + 8\), for some integer \(m\).

Solution:

Based on Euclid's Division Lemma, any positive integer (n) can be expressed in one of the forms when divided by 3 as:

\[ n = 3q + r \quad \, where \; q \in Z, \; and \; 0 \le r < 3 \]

That is, \(n\) can take the form:

• \(n = 3q\)
• \(n = 3q + 1\)
• \(n = 3q + 2\)

Next, let's compute \(n^3\) for each possible value of \(n\):

Case 1: \(n = 3q\)

\[ n^3 = (3q)^3 = 27q^3 = 9(3q^3) \] Thus, \( n^3 \) is divisible by 9, meaning \( n^3 = 9m \) for some integer \( m \). .

Case 2: \(n = 3q + 1\)

\[ n^3 = (3q + 1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1 \] So, \(n^3 = 9m + 1\)

Case 3: \(n = 3q + 2\)

\[ n^3 = (3q + 2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8 \] So, \(n^3 = 9m + 8\)

In all three cases, the cube of any positive integer is of the form:

• \(9m\)
• \(9m + 1\)
• \(9m + 8\)

Hence proved using Euclid’s Division Lemma.